On Diophantine Equations 2^x ± (2^kp)^y = z² and −2^x + (2^k3)^y = z²

Abstract

In this paper, we solve three Diophantine equations: $2^x\pm {\left(2^{k}p\right)}^y=z^2$ and $-2^x+{\left(2^{k}3\right)}^y=z^2$ with $k\ge 0$ and prime $p\equiv \pm 3$$(mod8)$. We obtain all the non-negative integer solutions by using elementary methods and the database of elliptic curves in “The L-functions and modular forms database” (LMFDB).

1. Introduction

For various fixed pairs of integers $(a,b)$, the exponential Diophantine equations $a^x+b^y=z^2$ and $a^x-b^y=z^2$ have received a lot of attention in recent decades [1,2,3,4,5,6]. A summary of the most recent works on the equation $a^x+b^y=z^2$ can be found in [2,4]. One can find a comprehensive study of exponential Diophantine equations in [7].

In Section 2, we list some results needed in the proof of our main theorems. In Section 3, we solve three Diophantine equations: $2^x\pm {\left(2^{k}p\right)}^y=z^2$ and $-2^x+{\left(2^{k}3\right)}^y=z^2$ with non-negative integer k and prime $p\equiv \pm 3$$(mod8)$. The author of [8] first solved the equation $2^x+3^y=z^2$. When $p=5$, the equation $2^x\pm {\left(2^{k}p\right)}^y=z^2$ was solved in [9]. When $p=3$ and $k=0,1$, the equation $2^x\pm {\left(2^{k}p\right)}^y=z^2$ was solved in [10]. We could not generally solve the equation $-2^x+{\left(2^{k}p\right)}^y=z^2$ since we need to use the database of [11]. But for $p=5$, it was solved in [9]. For $p=3$ and $k=0,1$, it was also solved in [10]. Our works in this paper can be taken as a generalization of some results in [8,9,10]. We point out that our methods used in this paper can be applied to the equation $-2^x+{\left(2^{k}p\right)}^y=z^2$ for any specific value of prime p with $p\equiv 3,5,7(mod8)$ with the help of database LMFDB [11]. “elliptic curve cryptography” (ECC) is a type of public-key cryptography that utilizes the mathematical properties of elliptic curves. A future study of the equation $-2^x+{\left(2^{k}p\right)}^y=z^2$ with general p may have some potential applications in cryptography since these equations are related to elliptic curves.

2. Preliminaries

The following lemma was conjectured by Eugène Charles Catalan in 1844 and proved in 2002 by Preda Mihăilescu [12].

Lemma 1

(Catalan–Mihăilescu Theorem [12]). If a, b, x, and y are integers and $\min \{a,b,x,y\}$ $>1$, then the Diophantine equation $a^x-b^y=1$ has a unique integer solution $(a,b,x,y)=(3,2,2,3)$.

In 2003, Leu and Li proved the following result [13].

Lemma 2

([13]). The Diophantine equation $2x^2+1=3^y$ has exactly three positive rational integral solutions, namely $(x,y)=(1,1),(2,2)$, and $(11,5)$.

The discriminant of the elliptic curve $y^2=x^3+ax+b$ over $\mathbb{Q}$ is $\Delta =-16(4a^3+27b^2)$.

Using discriminant, the following lemma can be found in the database of elliptic curves in “The L-functions and modular forms database” (LMFDB) [11].

Lemma 3

([11]). The three elliptic curves over $\mathbb{Q}$ and their integer solutions are listed below.

(1) $y^2=x^3-2$ has only two integer solutions $(x,y)=(3,\pm 5)$.

(2) $y^2=x^3-18$ has only two integer solutions $(x,y)=(3,\pm 3)$.

(3) $y^2=x^3-162$ has no integer solutions.

3. Solutions of the Equations ${\mathbf{2}}^{\mathit{x}}\pm {\left({\mathbf{2}}^{\mathit{k}}\mathit{p}\right)}^{\mathit{y}}={\mathit{z}}^{\mathbf{2}}$ and $-{\mathbf{2}}^{\mathit{x}}+{\left({\mathbf{2}}^{\mathit{k}}\mathbf{3}\right)}^{\mathit{y}}={\mathit{z}}^{\mathbf{2}}$

In this section, we solve three Diophantine equations $2^x\pm {\left(2^{k}p\right)}^y=z^2$ and $-2^x+{\left(2^{k}3\right)}^y=z^2$ with $k\ge 0$ and prime $p\equiv \pm 3(mod8)$.

In the proofs of the theorems, we frequently use the following facts which can be found in any standard elementary textbook.

(1): Let p be an odd prime and $\left({\displaystyle \frac{a}{p}}\right)$ the Legendre symbol. We have $\left({\displaystyle \frac{2}{p}}\right)=1$ if and only if $p\equiv \pm 1$ (mod 8), or equivalently, $\left({\displaystyle \frac{2}{p}}\right)=-1$ if and only if $p\equiv \pm 3$ (mod 8).

(2): For any odd number a, we have $a^2\equiv 1$ (mod 8).

Theorem 1.

For $k\ge 0$ and prime $p\equiv \pm 3$ (mod 8), the Diophantine equation

$$2^x+{\left(2^{k}p\right)}^y=z^2$$

(1)

has only the following non-negative integer solutions:

1.

$(p,k,x,y,z)=(p,k,3,0,3)$;

2.

$(p,k,x,y,z)=(3,2n-2,2n-2,1,2^n)$;

3.

$(p,k,x,y,z)=(3,2n+2,2n-2,1,7·2^{n-1})$;

4.

$(p,k,x,y,z)=(3,2n+1,2n-2,1,5·2^{n-1})$;

5.

$(p,k,x,y,z)=(3,n-1,2n+2,2,5·2^{n-1})$;

6.

$(p,k,x,y,z)=(5,2n-2,2n,1,3·2^{n-1})$;

7.

$(p,k,x,y,z)=(5,2n+2,2n-2,1,9·2^{n-1})$,

where n is any positive integer.

Proof. 

If $y=0$, then $(p,k,x,y,z)=(p,k,3,0,3)$ by Lemma 1.

For $y\ge 1$, through Equation (1), we have $2^x\equiv z^2(modp)$. Hence, ${\left({\displaystyle \frac{2}{p}}\right)}^x={(-1)}^x=\left({\displaystyle \frac{z^2}{p}}\right)=1$ since $p\equiv \pm 3(mod8)$. Therefore, we obtain $x=2m$ for some non-negative integer m. We write Equation (1) as

$$(z-2^m)(z+2^m)=2^{ky}{p}^y.$$

Letting $d=gcd(z-2^m,z+2^m)$, we have $d\mid (z+2^m)-(z-2^m)=2^{m+1}$. Hence, $d=2^t$, $0\le t\le m+1$. We also have $d^2\mid (z+2^m)(z-2^m)=2^{ky}{p}^y$, so, $0\le 2t\le ky$. In summary, we obtain $0\le t\le min\{m+1,\lfloor {\displaystyle \frac{ky}{2}}\rfloor \}$.

We have the following four possibilities, namely, $z-2^m=2^t,2^t{p}^y,2^{ky-t},2^{ky-t}{p}^y$.

Case 1. $z-2^m=2^t$ and $z+2^m=2^{ky-t}{p}^y$.

In this case, we have $2^{m+1}=2^{ky-t}{p}^y-2^t$. Hence, $2^{m+1-t}=2^{ky-2t}{p}^y-1$. It is clear that $m+1-t=0$ is impossible since $y\ge 1$. For $m+1-t\ge 1$, $2^{m+1-t}$ is even. So is $2^{ky-2t}{p}^y-1$. Therefore, $ky-2t=0$. So, we obtain $2^{m+1-t}=p^y-1$ or $p^y-2^{m+1-t}=1$.

Case 1.1. $y=1$.

We have $p=2^{m+1-t}+1$. Since $p\equiv \pm 3(mod8)$, we obtain $1\le m+1-t\le 2$.

Case 1.1.1. $m+1-t=1$.

We have $m=t$, $x=2t$. From $ky-2t=0$, we have $k=2t$, $z=2^m+2^t=2^{t+1}$. Letting $n=t+1$, we have $(p,k,x,y,z)=(3,2n-2,2n-2,1,2^n)$.

Case 1.1.2. $m+1-t=2$.

We have $m=t+1$, $x=2t+2$, $k=2t$, and $z=2^m+2^t=3·2^t$. Letting $n=t+1$, we have $(p,k,x,y,z)=(5,2n-2,2n,1,3·2^{n-1})$.

Case 1.2. $y\ge 2$.

From $p^y-2^{m+1-t}=1$, we obtain $2^{m+1-t}=p^y-1\ge 3^2-1=8$. Hence, we have $m+1-t\ge 3$ and $min\{p,2,y,m+1-t\}>1$, we obtain $p=3$, $y=2$ and $m+1-t=3$ by Lemma 1. Hence, $m=t+2$, $x=2t+4$, $z=2^m+2^t=5·2^t$ and $k=t$. Letting $n=t+1$, we obtain $(p,k,x,y,z)=(3,n-1,2n+2,2,5·2^{n-1})$

Case 2. $z-2^m=2^t{p}^y$ and $z+2^m=2^{ky-t}$.

In this case, we have $2^{m+1}=2^{ky-t}-2^t{p}^y$, so, $2^{m+1-t}=2^{ky-2t}-p^y$. If $m+1-t\ge 1$, then $ky-2t=0$ since both sides must be even. We obtain $2^{m+1-t}+p^y=1$, which is impossible. Hence, we obtain $m+1-t=0$ and $1=2^{ky-2t}-p^y$. Since $y\ge 1$, by Lemma 1, we know $y=1$ and $p=2^{ky-2t}-1$. If $ky-2t\ge 3$, we obtain $p\equiv -1(mod8)$ which is contrary to $p\equiv \pm 3(mod8)$. So, $ky-2t=2$, $p=3$, $k=2t+2$, $m=t-1$, $x=2t-2$ and $z=2^m+2^t{p}^y=2^{t-1}+3·2^t=7·2^{t-1}$. Letting $n=t$ since $t\ge 1$, we have $(p,k,x,y,z)=(3,2n+2,2n-2,1,7·2^{n-1})$.

Case 3. $z-2^m=2^{ky-t}$ and $z+2^m=2^t{p}^y$.

In this case, we have $2^{m+1}=2^t{p}^y-2^{ky-t}$, so, $2^{m+1-t}=p^y-2^{ky-2t}$.

Case 3.1. $m+1-t=0$.

We have $p^y-2^{ky-2t}=1$.

Case 3.1.1. $y=1$.

Since $p=2^{ky-2t}+1\equiv \pm 3(mod8)$, we obtain $1\le ky-2t\le 2$, there are two possibilities.

Case 3.1.1.1. $ky-2t=1$.

We have $m=t-1$, $y=1$, $k=2t+1$, $p=3$, $x=2t-2$ and $z=2^m+2^{ky-t}=2^{t-1}+2^{t+1}=5·2^{t-1}$. Letting $n=t$ since $t\ge 1$, we obtain $(p,k,x,y,z)=(3,2n+1,2n-2,1,5·2^{n-1})$.

Case 3.1.1.2. $ky-2t=2$.

We have $m=t-1$, $y=1$, $k=2t+2$, $p=5$, $x=2t-2$ and $z=2^m+2^{ky-t}=2^{t-1}+2^{t+2}=9·2^{t-1}$. Letting $n=t$ since $t\ge 1$, we obtain $(p,k,x,y,z)=(5,2n+2,2n-2,1,9·2^{n-1})$.

Case 3.1.2. $y\ge 2$.

From $p^y-2^{ky-2t}=1$, since $y\ge 2$ and $p\ge 3$, we obtain $ky-2t\ge 2$. Hence, $min\{p,2,y,ky-2t\}>1$. By Lemma 1, we obtain $(p,2,y,ky-2t)=(3,2,2,3)$. Namely, $ky-2t=2k-2t=3$ which is impossible. So, there is no solution in this case.

Case 3.2. $m+1-t\ge 1$.

From $2^{m+1-t}=p^y-2^{ky-2t}$, we know $ky-2t=0$ since both sides are even. We obtain $p^y-2^{m+1-t}=1$.

Case 3.2.1. $y=1$.

We have $p=2^{m+1-t}+1\equiv \pm 3(mod8)$. Hence, $m+1-t=1,2$.

Case 3.2.1.1. $m+1-t=1$.

We obtain $m=t$, $p=3$, $y=1$, $x=2t$, $ky=k=2t$ and $z=2^m+2^{ky-t}=2^t+2^t=2^{t+1}$. Letting $n=t+1$, we have $(p,k,x,y,z)=(3,2n-2,2n-2,1,2^n)$ again (We obtain this solution first in Case 1.1.1.).

Case 3.2.1.2. $m+1-t=2$.

We obtain $m=t+1$, $p=5$, $y=1$, $x=2t+2$, $ky=k=2t$ and $z=2^m+2^{ky-t}=2^{t+1}+2^t=3·2^t$. Letting $n=t+1$, we have $(p,k,x,y,z)=(5,2n-2,2n,1,3·2^{n-1})$ again (We obtain this solution first in Case 1.1.2.).

Case 3.2.2. $y\ge 2$.

In $p^y-2^{m+1-t}=1$, from $y\ge 2$, we obtain $m+1-t\ge 2$. Hence, $min\{p,2,y,m+1-t\}>1$. By Lemma 1, we know $p=3$, $y=2$, $m+1-t=3$, $ky=2k=2t$. So, $m=t+2$, $x=2t+4$, $z=2^{t+2}+2^t=5·2^t$. Letting $n=t+1$, we obtain $(p,k,x,y,z)=(3,n-1,2n+2,2,5·2^{n-1})$ again (we obtain this solution first in Case 1.2.).

Case 4. $z-2^m=2^{ky-t}{p}^y$ and $z+2^m=2^t$.

In this case, we have $2^{m+1}=2^t-2^{ky-t}{p}^y$. Hence, we have $2^{m+1-t}+2^{ky-2t}{p}^y=1$, which is impossible. □

Theorem 2.

For $k\ge 0$ and prime $p\equiv \pm 3(mod8)$, the Diophantine equation

$$2^x-{\left(2^{k}p\right)}^y=z^2$$

(2)

has only the following non-negative integer solutions:

1.

$(p,k,x,y,z)=(p,k,0,0,0)$;

2.

$(p,k,x,y,z)=(p,k,1,0,1)$;

3.

$(p,k,x,y,z)=(3,2n-2,2n,1,2^{n-1})$,

where n is any positive integer.

Proof. 

For $x=0$, we have $(p,k,x,y,z)=(p,k,0,0,0)$.

For $x=1$, we have $(p,k,x,y,z)=(p,k,1,0,1)$.

From now on, we have $x\ge 2$. If $y=0$, then we obtain $-1\equiv z^2(mod4)$, which is absurd. Hence, we obtain $y\ge 1$. Taking module p in Equation (2), we obtain $2^x\equiv z^2(modp)$. So $x=2m$ for some integers, $m\ge 1$, since $\left({\displaystyle \frac{2}{p}}\right)=-1$.

Equation (2) can be written as

$$(2^m-z)(2^m+z)=2^{ky}{p}^y.$$

Letting $d=gcd(2^m-z,2^m+z)$, we have $d\mid 2^{m+1}$ and $d^2\mid 2^{ky}{p}^y$.

Hence, $d=2^t$, $0\le t\le min\{m+1,\lfloor {\displaystyle \frac{ky}{2}}\rfloor \}$.

There are four cases, namely, $2^m-z=2^t,2^t{p}^y,2^{ky-t},2^{ky-t}{p}^y$.

Case 1. $2^m-z=2^t$ and $2^m+z=2^{ky-t}{p}^y$.

We have $2^{m+1}=2^{ky-t}{p}^y+2^t$, or $2^{m+1-t}=2^{ky-2t}{p}^y+1$.

It is clear that $m+1-t\ge 1$ and $2^{m+1-t}$ is even. Hence, $2^{ky-2t}{p}^y+1$ is also even. So, we obtain $ky=2t$.

Now we have $2^{m+1-t}=p^y+1$, or $2^{m+1-t}-p^y=1$. By Lemma 1, we obtain $y=1$ and $p=2^{m+1-t}-1$. Because $p\equiv \pm 3(mod8)$, we know $m+1-t=2$. After simple calculation, we obtain $(p,k,x,y,z)=(3,2t,2t+2,1,2^t)$. Letting $n=t+1$, we obtain $(p,k,x,y,z)=(3,2n-2,2n,1,2^{n-1})$.

Case 2. $2^m-z=2^t{p}^y$ and $2^m+z=2^{ky-t}$.

We have $2^{m+1}=2^t{p}^y+2^{ky-t}$, or $2^{m+1-t}=p^y+2^{ky-2t}$.

It is clear that $m+1-t\ge 1$. So, we obtain $ky-2t=0$ since both sides are even. Then $2^m-z=2^t{p}^y>2^{ky-t}=2^m+z$, a contradiction. Hence, there is no solution in this case.

Case 3. $2^m-z=2^{ky-t}$ and $2^m+z=2^t{p}^y$.

We have $2^{m+1}=2^{ky-t}+2^t{p}^y$, or $2^{m+1-t}=2^{ky-2t}+p^y$.

It is clear that $m+1-t\ge 1$ and $2^{m+1-t}$ is even. Hence, $2^{ky-2t}+p^y$ is also even. So, we obtain $ky=2t$.

Now we have $2^{m+1-t}=p^y+1$, or $2^{m+1-t}-p^y=1$. By Lemma 1, we obtain $y=1$. Same as before, we have $m+1-t=2$. After simple calculation, we obtain $(p,k,x,y,z)=(3,2t,2t+2,1,2^t)$. Letting $n=t+1$, we obtain $(p,k,x,y,z)=(3,2n-2,2n,1,2^{n-1})$. This solution has been obtained in Case 1.

Case 4. $2^m-z=2^{ky-t}{p}^y$ and $2^m+z=2^t$.

Since $2^{ky-t}{p}^y>2^t$ due to $ky-t\ge t$ and $y\ge 1$, we obtain $2^m-z>2^m+z$, a contradiction. Hence, there is no non-negative integer solution in this case. □

Theorem 3.

For $k\ge 0$, the Diophantine equation

$$-2^x+{\left(2^{k}3\right)}^y=z^2$$

(3)

has only the following non-negative integer solutions:

1.

$(k,x,y,z)=(k,0,0,0)$;

2.

$(k,x,y,z)=(n-1,2n+1,2,2^{n-1})$;

3.

$(k,x,y,z)=(n-1,4n+1,4,7·2^{2n-2})$;

4.

$(k,x,y,z)=(2n-1,2n-1,1,2^n)$;

5.

$(k,x,y,z)=(2n-1,10n-5,5,88·2^{5n-5})$:

6.

$(k,x,y,z)=(2n-2,6n-5,3,5·2^{3n-3})$;

7.

$(k,x,y,z)=(2n-2,2n-1,1,2^{n-1})$,

where n is any positive integer.

Proof. 

If $y=0$, then $(k,x,y,z)=(k,0,0,0)$.

For $y\ge 1$, taking module 3 in Equation (3), we have ${(-1)}^{x+1}\equiv z^2(mod3)$. It is clear that this congruence holds only if $2\nmid x$.

There are two cases, namely, $2\mid y$ and $2\nmid y$.

Case 1. $2\nmid x$ and $2\mid y$.

Letting $y=2m$, $m\ge 1$, we obtain $(2^{km}{3}^m-z)(2^{km}{3}^m+z)=2^x$, which implies $2^{km}{3}^m-z=2^t$ and $2^{km}{3}^m+z=2^{x-t}$ with $0\le t<x-t$ since $z\ne 0$.

So we have

$$2^{km+1-t}{3}^m=2^{x-2t}+1.$$

Since the right side is odd, we obtain $km+1-t=0$ and $3^m-2^{x-2t}=1$. By Lemma 1, we receive $(m,x-2t)=(1,1),(2,3)$.

For $(m,x-2t)=(1,1)$, we obtain $m=1$, $x=2t+1$, $k=t-1$, $y=2$ and $z=2^{t-1}$. Letting $n=t$, we obtain a solution $(k,x,y,z)=(n-1,2n+1,2,2^{n-1})$.

For $(m,x-2t)=(2,3)$, we obtain $m=2$, $x=2t+3$, $2k+1=t$, $y=4$ and $z=7·2^{t-1}$. Hence, $(k,x,y,z)=(k,4k+5,4,7·2^{2k})$.

Letting $n=k+1$, we obtain a solution $(k,x,y,z)=(n-1,4n+1,4,7·2^{2n-2})$.

Case 2. $2\nmid x$ and $2\nmid y$.

Let $z=2^t{z}_1$, where $z_1\in \mathbb{N}$, $t\ge 0$ and $2\nmid z_1$.

Case 2.1. $ky>x$.

From Equation (3), we obtain $-2^x+{\left(2^{k}3\right)}^y=2^{2t}{z}_1^2$, $2^x(-1+2^{ky-x}{3}^y)=2^{2t}{z}_1^2$. Hence, we obtain $x=2t$, which is contrary to $2\nmid x$.

Case 2.2. $ky=x$.

From Equation (3), we obtain $2^x(-1+3^y)=2^{2t}{z}_1^2$, or $2^{x+1}(3^{y-1}+\cdots +3+1)=2^{2t}{z}_1^2$. Since $2\nmid y$, we know $3^{y-1}+\cdots +3+1$ is odd. So, we obtain $x+1=2t$ and

$$3^y=2z_1^2+1.$$

By Lemma 2, we obtain $(y,z_1)=(1,1),(5,11)$.

For $(y,z_1)=(1,1)$, we obtain $y=1$, $x=2t-1$, $k=2t-1$, $z=2^t$. Letting $n=t$, we obtain the solution $(k,x,y,z)=(2n-1,2n-1,1,2^n)$.

For $(y,z_1)=(5,11)$, we obtain $x=2t-1$, $y=5$, $z=11·2^t$. From $ky=x$, we obtian $5k=2t-1$. Since k is odd, letting $k=2k_1+1$, we obtain $t=5k_1+3$. Letting $k_1=n-1$, we obtain the solution $(k,x,y,z)=(2n-1,10n-5,5,88·2^{5n-5})$.

Case 2.3. $ky<x$.

From Equation (3), we obtain $2^{ky}(-2^{x-ky}+3^y)=2^{2t}{z}_1^2$. Both $z_1$ and $-2^{x-ky}+3^y$ are odd. We obtain $ky=2t$ and $-2^{x-ky}+3^y=z_1^2$. If $x-ky\ge 2$, we obtain $z_1^2\equiv {(-1)}^y\equiv -1(mod4)$, which is absurd. So, we obtain $x-ky\le 1$. Because $ky=2t$ and x is odd, we obtain $x-ky=1$. Now, we only need to solve the following equation

$$z_1^2=3^y-2.$$

We consider three possibilities. Namely, $y=3s,3s+1,3s+2$.

Case 2.3.1. $y=3s$.

We have $z_1^2=3^{3s}-2={\left(3^s\right)}^3-2$, by Lemma 3, we obtain $(3^s,z_1)=(3,\pm 5)$. So, we obtain $s=1$, $z_1=5$, $y=3$, $3k=2t$, $x=yk+1=3k+1=2t+1$ and $z=5·2^t$. From $3k=2t$, we obtain $t=3t_1$ and $k=2t_1$. Letting $t_1=n-1$, we obtain a solution $(k,x,y,z)=(2n-2,6n-5,3,5·2^{3n-3})$.

Case 2.3.2. $y=3s+1$.

We have $z_1^2=3^{3s+1}-2$, or ${\left(3z_1\right)}^2={\left(3^{s+1}\right)}^3-18$; by Lemma 3, we obtain $(3^{s+1},3z_1)=(3,\pm 3)$. Hence, we have $s=0$, $z_1=1$, $y=1$, $z=2^t$, $k=2t$ and $x=ky+1=2t+1$. Letting $t=n-1$, we obtain a solution $(k,x,y,z)=(2n-2,2n-1,1,2^{n-1})$.

Case 2.3.3. $y=3s+2$.

We have $z_1^2=3^{3s+2}-2$, or ${\left(3^2{z}_1\right)}^2={\left(3^{s+2}\right)}^3-162$ by Lemma 3; there is no non-negative integer solution. □